1. The problem describes an angle formed by two rays sharing a common endpoint, with one ray horizontal to the right and the other ray diagonally upward to the right. 2. To find the measure of this angle, we consider the rays as vectors: the horizontal ray along the positive x-axis and the diagonal ray making some angle $\theta$ with the horizontal. 3. The horizontal ray can be represented as vector $\vec{u} = (1,0)$. 4. The diagonal ray can be represented as vector $\vec{v} = (x,y)$ where $x>0$ and $y>0$ since it points upward and to the right. 5. The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by the formula: $$\cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|}$$ 6. Since $\vec{u} = (1,0)$, the dot product $\vec{u} \cdot \vec{v} = 1 \cdot x + 0 \cdot y = x$. 7. The magnitude $\|\vec{u}\| = \sqrt{1^2 + 0^2} = 1$. 8. The magnitude $\|\vec{v}\| = \sqrt{x^2 + y^2}$. 9. Substitute these into the formula: $$\cos(\theta) = \frac{x}{1 \cdot \sqrt{x^2 + y^2}} = \frac{x}{\sqrt{x^2 + y^2}}$$ 10. To find $\theta$, take the inverse cosine: $$\theta = \cos^{-1}\left(\frac{x}{\sqrt{x^2 + y^2}}\right)$$ This formula allows you to calculate the angle between the horizontal ray and the diagonal ray given the coordinates of the diagonal ray's direction vector. Final answer: The angle $\theta$ formed is $$\theta = \cos^{-1}\left(\frac{x}{\sqrt{x^2 + y^2}}\right)$$ where $x$ and $y$ are the components of the diagonal ray's vector.